Given Problem:
$f$ is a function that satisfies the 3 following properties:
- $f:\mathbb{N}\rightarrow\mathbb{N}$
- $\sqrt{f(x)}\ge\frac{f(x)+f(1)}{2}$ for some x within the given domain
- $\frac{f(n)}{f(1)} = 2n - (f(1))^2 , n\ge2$
It is required to find all such functions.
Clarification: Consider that $0\notin\Bbb{N}$
Given Solution:
The first condition states that the domain and codomain of $f$ is $\Bbb{N}$. So, AM-GM inequality can be applied here as all the numbers that we are working with are positive. Applying the Inequality, we can get: $$\sqrt{f(x)}\ge\frac{f(1)+f(x)}{2}\ge\sqrt{f(1)f(x)}$$
But since $f(1)\ge1$, we have $$\sqrt{f(x)}\le\sqrt{f(1)f(x)}$$ So, the inequalities given above are actually equalities. Hence, for the inequality to hold, we must have $f(1) = 1$
Substituting this in 3rd condition gives:$$f(n) = 2n-1, n\ge2$$
Note that $f(1) = 2\cdot1-1$
So, $$f(n)=2n-1, n\in\Bbb{N}$$
My Confusion:
The second property does not hold for $x=2$.
L.H.S. $=\sqrt{f(2)}$ $=\sqrt{2\cdot2-1}$ $=\sqrt{3}$ $=1.7320508...$
R.H.S. $=\frac{f(2)+f(1)}{2}$ $=\frac{3+1}{2}$ $=2$
So, we have L.H.S. $\lt$ R.H.S Which clearly violates the second property. Where is the mistake?
I suppose the answer is that no such function exists, if you notice your inequality with $f(1) \geq 1$ gives us
$$\sqrt{f(x)}\ge\frac{f(1)+f(x)}{2}\ge\sqrt{f(1)f(x)} \geq \sqrt{f(x)}$$ So, not only do we get $$f(1)f(x) = f(x)\implies f(1) = 1$$ We also get $$\sqrt{f(x)}=\frac{f(1)+f(x)}{2}=\frac{1+f(x)}{2} \tag{1}$$
Now, we have arrived at $$f(n) = 2n-1$$ Now, substituting that in $(1)$ we see $$\sqrt{2n-1}=\frac{2n-1 + 1}{2}=n \implies n^2 - 2n +1=0 \implies n=1$$
So, no such function exists